Does ax+by = (a+b)z has a solution?

Question

For any pair of integers x and y, can we always find two integers a and b such that the equation $ax+by=(a+b)z$, $z$$\in$ $Z$

Answer 1

Yes, it can always be done.

Let $x,y$ be given integers.

Then choosing any integer $b$, and letting $a=-2b$, we have $a+b=-b$, hence \begin{align*} &ax + by \qquad\qquad\qquad\;\; \\[4pt] =\;\,&-2bx + by\\[4pt] =\;\,&-b(2x-y)\\[4pt] =\;\,&(a+b)(2x-y)\\[4pt] \end{align*} More generally, if $t,k$ are any integers, then letting \begin{align*} a&=kt \qquad\qquad\qquad \\[4pt] b&=(1-k)t\\[4pt] \end{align*} we have $a+b=t$, hence \begin{align*} &ax + by\\[4pt] =\;\,&(kt)x + (1-k)ty\\[4pt] =\;\,&t(kx+(1-k)y)\\[4pt] =\;\,&(a+b)\bigl(kx+(1-k)y\bigr)\\[4pt] \end{align*}